Solve y' - 2y = 3e^(2x) Step by Step | The integrating factor

Step by step

The question

Question: solve y′ − 2y = 3e^(2x).

y′ − 2y = 3e^(2x)

1 Standard form

Linear means this form: y′ + P y = Q.

P and Q have only x.

linear: y′ + P(x) y = Q(x)

Here P = −2 and Q = 3e^(2x).

P = −2, Q = 3e^(2x)

2 The integrating factor

The integrating factor is e to the integral of P.

Multiplying by it makes the left side one derivative.

integrating factor: μ = e^(∫P dx)

First the integral of P: the integral of −2 is −2x.

∫ −2 dx = −2x

So the integrating factor is e^(−2x).

No + C needed here: any one factor works.

μ = e^(−2x)

3 Multiply both sides

Multiply both sides by e^(−2x).

Every term, on both sides.

e^(−2x)y′ − 2e^(−2x)y = 3

Careful!

The right side is multiplied by e^(−2x) too.

Both sides, always.

Careful: not (e^(−2x)y)′ = 3e^(2x) ✗

4 One derivative on the left

By the product rule, the derivative of e^(−2x) × y is exactly our left side.

(μy)′ = μy′ + μ′y

μ′ = −−2e^(−2x)

So the left side is just (e^(−2x)y)′: one derivative.

(e^(−2x)y)′ = 3

5 Integrate both sides

Integrate both sides.

The integral of a derivative is the function itself.

e^(−2x)y = ∫ 3 dx

On the right: 3x, and + C.

e^(−2x)y = 3x + C

6 Solve for y

Divide both sides by e^(−2x).

y = 3x + Ce^(−2x)

Divide each term.

Dividing by e to a power is the same as e to minus that power.

y = 3xe^(2x) + Ce^(2x)

C over e^(−2x) = Ce^(2x)

7 Check

Let's check: put y back in.

First y′ is 3e^(2x) + 6xe^(2x) + 2Ce^(2x).

y′ = 3e^(2x) + 6xe^(2x) + 2Ce^(2x)

Then y′ − 2y.

y′ − 2y = 3e^(2x) + 6xe^(2x) + 2Ce^(2x) − 2(3xe^(2x) + Ce^(2x))

The C parts cancel, and we get 3e^(2x): Q again.

Right!

= 3e^(2x)

Let's look back.

First: standard form.

Then: the integrating factor.

Then: multiply both sides.

Then: one derivative on the left.

Then: integrate both sides.

Then: solve for y.

Last: check.

Answer